Question

Trig Identities

prove
sinx+1 \ 1-sinx = (tanx+secx)^2

Answer 1

If your expression mean:

( sin x + 1 ) / ( 1 - sin x ) = ( tan x + sec x )²

then:

( 1 + sin x ) / ( 1 - sin x ) = ( tan x + sec x )²

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Remark:

In expression:

( 1 + sin x ) / ( 1 - sin x )

Multiply numerator and denominator by 1 + sin x
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( 1 + sin x ) ∙ ( 1 + sin x ) / ( 1 - sin x ) ∙ ( 1 + sin x ) = tan²x + 2 ∙ tan x ∙ sec x + sec²x

( 1 + sin x )² / ( 1 ∙ 1 - sin x ∙ 1 + sin x ∙ 1 - sin x ∙ sin x ) = tan²x + 2 tan x ∙ sec x + sec²x

( 1 + sin x )² / ( 1 - sin x + sin x - sin²x ) = tan²x + 2 tan x ∙ sec x + sec²x

( 1 + sin x )² / ( 1 - sin²x ) = tan²x + 2 tan x ∙ sec x + sec²x

( 1² + 2 ∙ 1 ∙ sin x + sin²x ) / cos²x = tan²x + 2 tan x ∙ sec x + sec²x

( 1 + 2 sin x + sin²x ) / cos²x = tan²x + 2 tan x ∙ sec x + sec²x

1 / cos²x + 2 sin x / cos²x + sin²x / cos²x = tan²x + 2 tan x ∙ sec x + sec²x

sec²x + 2 sin x / ( cos x ∙ cosx )+ tan²x = tan²x + 2 tan x ∙ sec x + sec²x

sec²x + 2 sin x ∙ 1 / ( cos x ∙ cosx )+ tan²x = tan²x + 2 tan x ∙ sec x + sec²x

sec²x + 2 ( sin x / cos x ) ∙ 1 / cosx + tan²x = tan²x + 2 tan x ∙ sec x + sec²x

sec²x + 2 tan x ∙ sec x + tan²x = tan²x + 2 tan x ∙ sec x + sec²x

tan²x + 2 tan x ∙ sec x + sec²x = tan²x + 2 tan x ∙ sec x + sec²x

Answer 2

I would do it this way:
LS = (sinx+1)/(1-sinx)
= (sinx+1)/(1-sinx) * (sinx+1)/(1+sinx)
= (sinx+1)^2 /(1 - sin^2 x)
= (sinx + 1 )^2 / ((1+sinx)(1-sinx))
= (1+sinx)/(1-sinx)

RS = ( tan x + sec x )²
= (sinx/cosx + 1/cosx)^2
= ( (sinx+1)/cosx)^2
= (sinx+1)^2/cos^2 x
= (sinx+1)^2/(1- sin^2x) = (sinx+1)(sinx+1)/((1-sinx)(1+sinx)
= (sinx+1)/(1-sinx)
= LS

Answer 3

I think it goes "SINE SINE SINE, + SINE SINNERS SINE." "SINE." I think I won!